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$\begin{aligned} x+\log{(x)} = y-1\\ y+\log{(y-1)} = z-1 \\ z+\log{(z-2)} = x+2 \\ \end{aligned}$

There is only one solution set $\left(x,y,z\right)$ to the system of equations above. Evaluate $x+y+z$.

Note: Take the base of the logarithms as 10.

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